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The directional derivatives of $f$ at the origin in the directions of $\overrightarrow v =\langle1,-1\rangle$ and $\overrightarrow w =\langle\sqrt{3}, 1\rangle$ are $-\sqrt{2}$ and $4 + 3\sqrt{3}$ respectively. Find the maximum rate of change of $f$ at the origin.

Can someone help me answer? Or give me an idea or the steps I should do to answer this?
Thank you!

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Where is the function $f$? Are you working on the problem without it? – S. Snape Nov 24 '12 at 9:19
It wasn't given in the problem. – Rose Nov 24 '12 at 9:21
up vote 2 down vote accepted

The direction derivate along the vector $\langle a,b \rangle$ at origin is given as $$\langle a,b \rangle \cdot \langle f_x,f_y \rangle_{(0,0)} = af_x(0,0) + bf_y(0,0)$$ The direction derivate along the direction $\langle a,b \rangle$ at origin is given as $$\dfrac{\langle a,b \rangle}{\Vert \langle a,b \rangle \Vert} \cdot \langle f_x,f_y \rangle_{(0,0)} = \dfrac{af_x(0,0) + bf_y(0,0)}{\langle a,b \rangle}$$ You are given that $$\dfrac{f_x(0,0) - f_y(0,0)}{\sqrt{2}} = -\sqrt{2}$$ and $$\dfrac{\sqrt{3} f_x(0,0) + f_y(0,0)}2 = 4 + 3\sqrt{3}$$ Solving for $f_x(0,0)$ and $f_y(0,0)$ gives us $$f_x(0,0) = 6; \,\,\,\, f_y(0,0) = 8$$ The maximum rate of change is along $\vec{\nabla} f$ i.e. along the direction $\dfrac{\langle f_x,f_y \rangle}{\Vert \langle f_x,f_y \rangle \Vert}$. Hence, the maximum rate of change at the origin is $$\langle f_x,f_y \rangle \cdot \dfrac{\langle f_x,f_y \rangle}{\Vert \langle f_x,f_y \rangle \Vert} = \Vert \langle f_x,f_y \rangle \Vert = \sqrt{f_x(0,0)^2 + f_y(0,0)^2}$$ which gives us the answer as $10$.

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Wow. Thank you! A great help! :D – Rose Nov 24 '12 at 9:43
Wait, I realized something. Vectors v and w aren't unit vectors yet, do we need to get their equivalent unit vectors before we insert it into the two sets of equations? – Rose Nov 25 '12 at 7:09
@Rose It depends. If the question says that the gradient along a vector $v$ then it is $\vec{v} \cdot \vec{\nabla} f$. If it says the gradient along the direction of vector $v$, then it is $\dfrac{\vec{v}}{\Vert \vec{v} \Vert} \cdot \vec{\nabla} f$ – user17762 Nov 25 '12 at 7:11
@Rose I have edited the answer accordingly. – user17762 Nov 25 '12 at 7:15
Oh I see. Thanks! – Rose Nov 25 '12 at 10:27

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